# When a semigroup ideal is a determinantal ideal?

Let $$S=\langle n_1,...,n_r \rangle$$ be a commutative semigroup, and let $$I_S \subset k[x_1,...,x_r]$$ the associated ideal of $$S$$, defined as the kernel of the polinomial map $$\varphi:k[x_1,...x_n] \rightarrow k[t]$$, defined by $$\varphi(x_i)=t^{n_i}$$ for all $$i = 1,2,...,r$$. It is known that for $$r=3$$ the ideal $$I_S$$ is a complete intersection or is generated by the 2x2 minors of a 2x3 matrix like $$\begin{pmatrix} {x_1}^{a} & {x_2}^{b} & {x_3}^{c}\\ {x_2}^{d} & {x_3}^{e} & {x_1}^{f}\\ \end{pmatrix}$$ and so $$I_S$$, if is not a complete intersection, is a determinantal ideal. So I would ask:

1. There are some conditions to establish when $$I_S$$ is a determinantal ideal for $$r\geq4$$ ?
2. Is it possible generalize the case $$r=3$$? And so it could appens that for $$r\geq4$$, $$I_S$$ is the ideal generated by the 2x2 minors of a (r-1) x r matrix (or in general a (r-1) x m matrix)?
• For a general $(r-1)\times r$ matrix $A$ with entries in $k[x_{1},\ldots ,x_{r}]$, the locus where $\operatorname{rk}A\leq 1$ has codimension $(r-1)(r-2)$ in $\mathbb{A}^r$. Therefore it is indeed a curve for $r=3$, but it is empty for $r\geq 4$. So the answer to 2. is very probably no — unless you find a very particular matrix $A$. – abx Jan 31 at 15:32